Problem #1: In radical chlorination of alkanes, non-equivalent hydrogens react w

Question

Problem #1:

In radical chlorination of alkanes, non-equivalent hydrogens react with chlorine atoms at different rates. At 35 C, primary, secondary, and tertiary C-H bonds react at relative rates of 1 : 3.9 : 5.2 respectively.

These are conditions of kinetic control where product ratios are determined by relative rates of formation. For example, if A is formed twice as fast as B, the A:B product ratio will be 2.

Consider chlorination of the alkane below at 35 C.

1) Specify the most reactive C-H bond(s), a-c.
Two non-equivalent C-H bonds of comparable reactivity should be separated by commas, i.e. a,c.


2) Specify the site of chlorination in the major monochloro substitution product, a-c.
Two products that form in comparable quantities should be separated by commas, i.e. a,c

Problem#2:

Explanation / Answer

Answer – A) We are given the free radical chlorination of alkanes. We know the tertiary C-H bond is more reactive towards the free radical reaction than secondary C-H bond. The secondary C-H bond is more reactive towards the free radical reaction than primary C-H bond.

1)In this one we are given the 3,3-dimethylpentane and in this there is given a, b and c C-H bonds. The a is primary C-H bond, b is a secondary C-H bond and c is a primary C-H bond. The secondary C-H bond means b is the most reactive and a and c both are primary C-H bonds, so they both reacted with same rates.

So the most reactive C-H bond(s) is b , a=c.

2) In this one there are site of chlorination in the major monochloro substitution product is as follow – It is the same ranks as the reactivity of free radical, so the rank is- b, a=c

B) In this one we need to determine the how many monochloro substitution products are produced when the alkanes below are chlorinated and we need to consider constitutional isomers only, means on each different C there is only one -Cl will be attached.

a) In this one there are total 8 carbons, but out of the 8 carbons there are 4 pair of the carbons are same like the methyl at the quaternary C- there are 3 methyl on the same C, so showing same product and all these there methyl gives one product only, then tertiary C on the cyclobutane ring. Then the both secondary C on the cyclobutane which is 2nd C from the t-butyl group attached on the cyclobutane ring gives same product. The last one the 3rd C from the t-butyl group attached on the cyclobutane ring gives one product. So total there are 4 monochloro substitution products for this one alkane.

b) In this one we are given the heptane and there is plane of symmetrical at carbon number 4, so both side are same on each remaining 3 C and the 4th C is one, so total there are 4 monochloro substitution products for this one alkane

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